What is inverse oflaplace 2/(sin^2(+2)) ?

The answer to inverse oflaplace 2/(sin^2(+2)) is (2\delta(t))/(sin^2(2))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

inverse oflaplace 2/(sin^2(+2))

inverse laplace

\frac{2}{sin ^{2} ( + 2 )}

\frac{2 δ ( t )}{sin ^{2} ( 2 )}

Solution steps

L^{- 1} {\frac{2}{sin ^{2} ( 2 )}}

= L^{- 1} {\frac{2}{sin ^{2} ( 2 )} \cdot 1}

Use the constant multiplication property of Inverse Laplace Transform:

For function

f (t)

and constant

a : L^{- 1} {a \cdot f (t)} = a \cdot L^{- 1} {f (t)}

= \frac{2}{sin ^{2} ( 2 )} L^{- 1} {1}

Use Inverse Laplace Transform table:

L^{- 1} {1} = δ (t)

Where

δ (t)

is Dirac delta function

= \frac{2}{sin ^{2} ( 2 )} δ (t)

= \frac{2 δ ( t )}{sin ^{2} ( 2 )}

y^{^{''}} + y = sec (x) tan (x)

f (x) = x

x \to 0 lim (cos (π x))

\frac{\partial}{\partial x} (6 y)

\int e^{- 2 x} x^{2} d x

What is inverse oflaplace 2/(sin^2(+2)) ?
The answer to inverse oflaplace 2/(sin^2(+2)) is (2\delta(t))/(sin^2(2))